Comprehensive Notes on Mole Concept, Stoichiometric Yields, and Reagents

Concept of Mole and Molar Quantities

  • Definition of Mole:

    • A mole is defined as a collection of particles such as molecules, atoms, ions, neutrons, protons, etc., which is numerically equal to the number of atoms present in 12 gm12\,\text{gm} of the C-12\text{C-12} isotope.
    • 12 gm12\,\text{gm} of C-12\text{C-12} isotope (1 mole1\,\text{mole} of carbon) contains 6.023×10236.023 \times 10^{23} particles. This value is known as Avogadro's number and is denoted by NAN_A.

  • Mole in Terms of Mass (Molar Mass):

    • The total mass of one mole of any substance is known as its molar mass.
    • Molar mass is expressed in grams (gm\text{gm}).
    • Example: One mole of NaOH=40 gm\text{NaOH} = 40\,\text{gm}.

  • Mole in Terms of Number (Molar Number):

    • The total number of particles present in one mole of any substance is known as the molar number.
    • One mole of any substance always contains 6.023×10236.023 \times 10^{23} particles.

  • Mole in Terms of Volume (Molar Volume):

    • The volume occupied by one mole of any gaseous substance at Normal Temperature and Pressure (NTP) or Standard Temperature and Pressure (STP) is known as molar volume.
    • 1 mole1\,\text{mole} of any gaseous substance occupies 22.4 l22.4\,\text{l} at NTP.

Limiting and Excess Reagents

  • Limiting Reagent:

    • Definition: The chemical reagent which is present in an insufficient amount and is consumed completely first during a chemical reaction is known as the limiting reagent.
    • Role: It controls or limits the total amount of product formed.
    • Advantages:
    • Helps to calculate the total amount of product formed in a chemical reaction.
    • Allows calculation of the unreactive reagent leftover after reaction completion.
    • Assists in performing various stoichiometric chemical calculations.

  • Excess Reagent:

    • Definition: The chemical reagent which is present in a larger or greater amount and remains unreacted during or after the completion of the reaction is known as the excess reagent.
    • Role: It does not control or limit the total amount of product formed.

Chemical Yields: Theoretical, Experimental, and Percentage

  • Theoretical Yield:

    • The amount of product calculated from the complete reaction of the limiting reagent is known as theoretical yield.

  • Experimental Yield:

    • The amount of product obtained experimentally after the completion of a reaction is called experimental yield.

  • Percentage Yield:

    • The ratio of experimental yield to theoretical yield expressed as a percentage is known as percentage yield:

Percentage Yield=(Experimental YieldTheoretical Yield)×100%\text{Percentage Yield} = \left( \frac{\text{Experimental Yield}}{\text{Theoretical Yield}} \right) \times 100\%

  • Discrepancy Between Yields:
    • Theoretical yield is typically higher or greater than experimental yield because experimental processes may involve incomplete reactions, side reactions, material loss during transfer/isolation, or operational impurities.

Stoichiometric Problems and Practice Examples

  • Problem 1: Magnesium Content in Chlorophyll:

    • Question: Chlorophyll, a green coloring matter of plants responsible for photosynthesis, contains 2.68%2.68\% of Mg\text{Mg} by weight. Calculate the number of Mg\text{Mg} atoms in 2 gm2\,\text{gm} of chlorophyll.
    • Detailed Solution:
    • Mass of Mg\text{Mg} present = 2.68% of 2 gm=2.68100×2=0.0536 gm2.68\% \text{ of } 2\,\text{gm} = \frac{2.68}{100} \times 2 = 0.0536\,\text{gm}.
    • Moles of Mg\text{Mg} = 0.0536 gm24 g mol−1=0.002233 moles\frac{0.0536\,\text{gm}}{24\,\text{g\,mol}^{-1}} = 0.002233\,\text{moles}.
    • Number of Mg\text{Mg} atoms = 0.002233×6.023×1023=1.345×1021 atoms0.002233 \times 6.023 \times 10^{23} = 1.345 \times 10^{21}\text{ atoms}.

  • Problem 2: Gold Atoms in 20 Carat Gold:

    • Question: Calculate the number of gold atoms in 0.3 gm0.3\,\text{gm} of 20 carat20\,\text{carat} gold. Take 24 carat24\,\text{carat} gold as 100%100\% pure gold (assume atomic mass of gold = 197 amu197\,\text{amu}).
    • Detailed Solution:
    • Purity percentage = 2024×100%=83.33%\frac{20}{24} \times 100\% = 83.33\%.
    • Mass of pure gold = 0.3 gm×2024=0.25 gm0.3\,\text{gm} \times \frac{20}{24} = 0.25\,\text{gm}.
    • Moles of gold = 0.25 gm197 g mol−1=0.001269 moles\frac{0.25\,\text{gm}}{197\,\text{g\,mol}^{-1}} = 0.001269\,\text{moles}.
    • Number of gold atoms = 0.001269×6.023×1023=7.64×1020 atoms0.001269 \times 6.023 \times 10^{23} = 7.64 \times 10^{20}\text{ atoms}.

  • Problem 3: Total Electrons in Methane:

    • Question: Calculate the total number of electrons present in 1.6 gm1.6\,\text{gm} of methane (CH4\text{CH}_4).
    • Detailed Solution:
    • Molar mass of CH4=12+4(1)=16 g mol−1\text{CH}_4 = 12 + 4(1) = 16\,\text{g\,mol}^{-1}.
    • Moles of CH4=1.6 gm16 g mol−1=0.1 moles\text{CH}_4 = \frac{1.6\,\text{gm}}{16\,\text{g\,mol}^{-1}} = 0.1\,\text{moles}.
    • Molecules of CH4=0.1×6.023×1023=6.023×1022 molecules\text{CH}_4 = 0.1 \times 6.023 \times 10^{23} = 6.023 \times 10^{22}\text{ molecules}.
    • Electrons per CH4\text{CH}_4 molecule = 6 (from C)+4×1 (from H)=10 electrons6 \text{ (from C)} + 4 \times 1 \text{ (from H)} = 10\text{ electrons}.
    • Total number of electrons = 10×6.023×1022=6.023×1023 electrons10 \times 6.023 \times 10^{22} = 6.023 \times 10^{23}\text{ electrons}.

  • Problem 4: Water Molecules in a Spherical Drop:

    • Question: Calculate the number of water molecules present in a spherical drop of water having a radius of 1 mm1\,\text{mm} [given density of water = 1 g cm−31\,\text{g\,cm}^{-3}].
    • Detailed Solution:
    • Radius r=1 mm=0.1 cmr = 1\,\text{mm} = 0.1\,\text{cm}.
    • Volume of spherical drop V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times (0.1\,\text{cm})^3 = 0.004189\,\text{cm}^3$.\n * Mass of water drop = \text{Volume} \times \text{Density} = 0.004189\,\text{cm}^3 \times 1\,\text{g\,cm}^{-3} = 0.004189\,\text{gm}.\n * Moles of \text{H}_2\text{O} = \frac{0.004189\,\text{gm}}{18\,\text{g\,mol}^{-1}} = 0.0002327\,\text{moles}.\n * Number of water molecules = 0.0002327 \times 6.023 \times 10^{23} = 1.40 \times 10^{20}\text{ molecules}.\n\n* **Problem 5: Molecular Weight of a Gas from Volume at NTP**:\n * **Question**: What will be the molecular weight of a gas 11.2\,\text{l}ofwhichatNTPweightsof which at NTP weights14\,\text{gm}?\n * **Detailed Solution**:\n * Moles of gas at NTP = \frac{11.2\,\text{l}}{22.4\,\text{l\,mol}^{-1}} = 0.5\,\text{moles}.\n * Molecular weight = \frac{\text{Mass}}{\text{Moles}} = \frac{14\,\text{gm}}{0.5\,\text{moles}} = 28\,\text{g\,mol}^{-1}.\n\n* **Problem 6: Weight Comparison Between Sodium and Carbon**:\n * **Question**: What weight of sodium will contain the same number of atoms as are present in 1.2\,\text{gm} of Carbon?\n * **Detailed Solution**:\n * Moles of Carbon = \frac{1.2\,\text{gm}}{12\,\text{g\,mol}^{-1}} = 0.1\,\text{moles}.\n * Equal number of atoms implies equal number of moles, so required moles of Sodium (\text{Na})=) =0.1\,\text{moles}.\n * Atomic mass of \text{Na} = 23\,\text{g\,mol}^{-1}.\n * Required mass of \text{Na} = 0.1\,\text{moles} \times 23\,\text{g\,mol}^{-1} = 2.3\,\text{gm}.\n\n* **Problem 7: Burning Magnesium in Oxygen**:\n * **Question**: 1\,\text{gm}ofof\text{Mg}isburnedinaclosedvesselwhichcontainsis burned in a closed vessel which contains0.5\,\text{gm}ofof\text{O}_2. Find:\n * a. Which one is limiting reagent and why?\n * b. Number of moles of excess reagent.\n * c. Mass, moles, and molecules of \text{MgO} formed.\n * **Detailed Solution**:\n * Balanced equation: 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}.\n * Stoichiometric mass ratio: 2 \times 24\,\text{gm} = 48\,\text{gm}ofof\text{Mg}reactswithreacts with32\,\text{gm}ofof\text{O}_2\n * Mass of \text{Mg}neededforneeded for0.5\,\text{gm}ofof\text{O}_2 = \frac{48}{32} \times 0.5\,\text{gm} = 0.75\,\text{gm}.\n * Since 1\,\text{gm}ofof\text{Mg}isprovidedandonlyis provided and only0.75\,\text{gm}isrequired,is required,\text{O}_2 is completely consumed first and is therefore the **limiting reagent**.\n * Mass of excess \text{Mg}leftover=leftover =1\,\text{gm} - 0.75\,\text{gm} = 0.25\,\text{gm}.\n * Moles of excess \text{Mg} = \frac{0.25\,\text{gm}}{24\,\text{g\,mol}^{-1}} = 0.01042\,\text{moles}.\n * Moles of \text{O}_2reacted=reacted =\frac{0.5\,\text{gm}}{32\,\text{g\,mol}^{-1}} = 0.015625\,\text{moles}.\n * Moles of \text{MgO}formed=formed =2 \times 0.015625 = 0.03125\,\text{moles}.\n * Mass of \text{MgO}formed=formed =0.03125\,\text{moles} \times 40\,\text{g\,mol}^{-1} = 1.25\,\text{gm}.\n * Molecules of \text{MgO}formed=formed =0.03125 \times 6.023 \times 10^{23} = 1.882 \times 10^{22}\text{ molecules}.\n\n* **Problem 8: Reaction of Calcium Hydroxide and Ammonium Chloride**:\n * **Question**: For the reaction:\n\n\text{Ca(OH)}_2 + 2\text{NH}_4\text{Cl} \rightarrow \text{CaCl}_2 + 2\text{NH}_3 + 2\text{H}_2\text{O}\n\n * The reaction is carried out by mixing 7\,\text{gm}ofpureof pure\text{Ca(OH)}_2andand\text{NH}_4\text{Cl}. Determine the limiting reagent and yield of products.\n\n* **Problem 9: Reaction of Calcium Carbonate and Hydrochloric Acid**:\n * **Question**: A chemical reaction was carried out by mixing \text{CaCO}_3andand4.518 \times 10^{23}moleculesofmolecules of\text{HCl}(or(or0.25\,\text{moles}ofof\text{HCl})togive) to give\text{CaCl}_2,,\text{H}_2\text{O},and, and\text{CO}_2 according to:\n\n\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\n\n * Calculate:\n * a. Limiting reagent and why.\n * b. Mass of \text{CaCl}_2 formed.\n * c. Molecules of excess reagent leftover in the reaction.\n * d. Molecules of \text{H}_2\text{O} formed.\n * e. Volume of \text{CO}_2gasproducedatgas produced at27^\circ\text{C}andand770\,\text{mm} pressure.\n\n* **Problem 10: Impure Marble with Concentrated Hydrochloric Acid Solution**:\n * **Question**: 1.5\,\text{gram}ofmarblecontainingof marble containing75\%ofof\text{CaCO}_3istreatedwithis treated with20\,\text{ml}ofof36\%purepure\text{HCl}solutionhavingspecificgravitysolution having specific gravity1.18. Calculate:\n * a. Limiting reagent and why.\n * b. Number of moles and molecules of excess reagent.\n * c. Mass of salt formed.\n * d. Volume of liberated \text{CO}_2 at NTP.\n * **Detailed Solution**:\n * Mass of pure \text{CaCO}_3 = \frac{75}{100} \times 1.5\,\text{gm} = 1.125\,\text{gm}.\n * Total mass of \text{HCl}solution=solution =\text{Volume} \times \text{Specific Gravity} = 20\,\text{ml} \times 1.18\,\text{g\,cm}^{-3} = 23.6\,\text{gm}.\n * Mass of pure \text{HCl} = \frac{36}{100} \times 23.6\,\text{gm} = 8.496\,\text{gm}.\n * Moles of \text{CaCO}_3 = \frac{1.125\,\text{gm}}{100\,\text{g\,mol}^{-1}} = 0.01125\,\text{moles}.\n * Moles of \text{HCl} = \frac{8.496\,\text{gm}}{36.5\,\text{g\,mol}^{-1}} = 0.23276\,\text{moles}.\n * Stoichiometric ratio requires 2\,\text{moles}ofof\text{HCl}perper1\,\text{mole}ofof\text{CaCO}_3.Molesof. Moles of\text{HCl}required=required =2 \times 0.01125 = 0.0225\,\text{moles}.\n * Since 0.23276\,\text{moles}ofof\text{HCl}arepresent,are present,\text{CaCO}_3 is completely consumed first and is the **limiting reagent**.\n * Excess \text{HCl}molesremaining=moles remaining =0.23276 - 0.0225 = 0.21026\,\text{moles}.\n * Excess \text{HCl}molecules=molecules =0.21026 \times 6.023 \times 10^{23} = 1.266 \times 10^{23}\text{ molecules}.\n * Mass of salt (\text{CaCl}_2)formed:Molesof) formed: Moles of\text{CaCl}_2 = 0.01125\,\text{moles}.Molarmassof. Molar mass of\text{CaCl}_2 = 111\,\text{g\,mol}^{-1}.Mass=. Mass =0.01125 \times 111 = 1.24875\,\text{gm}.\n * Volume of liberated \text{CO}_2atNTP:Molesofat NTP: Moles of\text{CO}_2 = 0.01125\,\text{moles}.Volume=. Volume =0.01125 \times 22.4\,\text{l} = 0.252\,\text{l}((252\,\text{ml}$$).